CODEWITHSUNDEEP
c++memory-managementpass-by-referencestdvectortemporary-objects
bigcodeszzer
bigcodeszzer

Reputation: 940

C++: return a std::vector<> member, by reference, to a temporary object

I have two functions.

function1 calls function2, which returns a reference to an object, that I then use to call a public function.

function1(function2).returned_object.mymethod();

This is a fairly unwieldy way to do things, but it works.

Now I have a problem where I need to call one method, then another method, so I need a temporary object.

ClassName temp_obj = function1(function2);

temp_obj.method1();
temp_obj.method2();

My question is, how can I declare a temporary object that will store the object that returns by reference. I'm guessing I need a pointer.

The other problem is, temp_obj is contained in a vector, and I'm concerned about conflicts there and memory leakage. If I use a pointer, will I then have to use a destructor also?

Here is the relevant code:

bool Machine::perform_function(std::vector<Job>& jobs) {
    bool add = true;

    if (can_fail) {
        add = coin_toss();

    }

    for (int i = 0; i < my_jobs.size(); i++) {

        if (add) {
            find_job(my_jobs[i], jobs).toggle(name, item_name, true);

        }
        if (!add) {
            find_job(my_jobs[i], jobs).toggle(name, item_name, false);

        }
    }

    return add;
}


Job& Machine::find_job(std::string jobname, std::vector<Job>& jobs) {

    for (int i = 0; i < jobs.size(); i++) {

        if (jobs[i].is_job(jobname)) {
            return jobs[i];

        }
    }
}

Upvotes: 0

Views: 134

Answers (4)

Peter
Peter

Reputation: 36597

Assuming it is function1() that returns a reference (your question is unclear on that), simply use a reference

 ClassName &temp_obj = function1(function2);

 temp_obj.method1();
 temp_obj.method2();

If function1() returns a const reference, than temp also needs to be const.

The limitation of this is that, once created, a reference cannot be reseated (made to refer to a different object). For example, in ;

 ClassName &temp_obj = function1(function2);    // initialisation

 temp_obj = function1(some_other_function);     // reassignment
 temp_obj.method1();
 temp_obj.method2();

the reassignment will not reseat temp, even if function1() returns a different reference on the second call. Instead, the object being referred to will be assigned (assuming it has a working assignment operator). If you want a variable that can refer to different objects over time, you need a pointer.

 ClassName *temp_obj = &function1(function2);    // assumes function1() returns a reference

 temp_obj = &function1(some_other_function);      //  reassigns the pointer

 temp_obj->method1();
 temp_obj->method2();

Upvotes: 1

TimDev
TimDev

Reputation: 95

Your question is unclear. But if you want to make a pointer to a function or member function there is a FAQ about such variables here.

https://isocpp.org/wiki/faq/pointers-to-members#fnptr-vs-memfnptr-types

Upvotes: -1

paddy
paddy

Reputation: 63471

It's perfectly fine to store the result as a reference:

ClassName & temp_obj = function1(function2);
temp_obj.method1();
temp_obj.method2();

Your other question here:

The other problem is, temp_obj is contained in a vector, and I'm concerned about conflicts there and memory leakage.

It's okay to return jobs[i] as a reference. The operator[] on a vector returns a reference itself. Provided you don't modify the vector in such a way as to change what that reference points at, and you are not storing a reference to an object that has been destroyed, you'll be fine to use it.

When your temporary reference goes out of scope, nothing happens. It's just a reference.

Upvotes: 1

Jack
Jack

Reputation: 133587

You need a temporary variable, but that temporary variable could be a reference so no copy is done and you are safe to use it.

But still better, you could use <algorithm> functions to do what you need without having to reinvent the wheel:

std::string name = ...;
auto it = std::find_if(begin(jobs), end(jobs), [&name] (const Job& job) { return job.is_job(name); });

if (it != jobs.end())
{
  it->method1();
  it->method2();
}

Upvotes: 2

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